= Solution
For each primitive linear, square or cubic <Bravais lattice> of spacing $a$, the <reciprocal lattice> has spacing $2\pi/a$. The first <Brillouin zone> is consequently the interval, square or cube with faces at components $k_i=\pm\pi/a$. The nearest zone boundary is a distance $k_B=\pi/a$ from the centre. The ratios are
$$
\boxed{\frac{k_F^{(1)}}{k_B}=\frac12,\qquad
\frac{k_F^{(2)}}{k_B}=\sqrt{\frac2\pi}=0.7979,\qquad
\frac{k_F^{(3)}}{k_B}=\left(\frac3\pi\right)^{1/3}=0.9847.}
$$
All are below one: the ideal free-electron <Fermi surface> lies wholly inside the first <Brillouin zone>. Its nearest approach to the boundary is in three dimensions, where it lies only about $1.53\%$ short of the nearest face. A lattice-induced <band gap> at a zone face therefore does not by itself make these half-filled systems insulating in the stated free-electron model. Interactions, structural changes and actual <electronic band structure> can alter that model prediction.
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